In this case, we know the sample means x^bar and sample variance s but we dont know the population variance. Then, the confidence interval formula is given by
[tex]\bar{x}-T_{n-1,\alpha/2}\frac{s}{\sqrt[]{n}}<\mu<\bar{x}+T_{n-1,\alpha/2}\frac{s}{\sqrt[]{n}}[/tex]
where n is the sample size of 46 and t_n-1,alpha/2 is the critical value of the t distribution with n-1=45 degrees of freedom.
From the t-table, for n-1=45 and alpha/2 = 0.025 (corresponding to 95% confidence level), we have that
[tex]T_{45,0.05/2}=2.014[/tex]
Then, by substituting this value and the remaining ones, we get
[tex]25.6-(2.014)\frac{6.1}{\sqrt[]{46}}<\mu<25.6+(2.014)\frac{6.1}{\sqrt[]{46}}[/tex]
which gives
[tex]25.6-1.8113<\mu<25.6+1.8113[/tex]
Then, by rounding to one decimal place, the answer is
[tex]23.8<\mu<27.4[/tex]
